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Question
Answer the following:
Let f: R → R be a function defined by f(x) = 5x3 – 8 for all x ∈ R, show that f is one-one and onto. Hence find f –1
Solution
f(x) = 5x3 – 8, x ∈ R
Let f(x1) = f(x2)
∴ 5x13 – 8 = 5x23 – 8
∴ x13 – x23 = 0
∴ `(x_1 - x_2) underbrace((x_1^2 + x_1 x_2 + x_2^2))_(> 0 "for all" x_1 * x_2 "as discriminant" < 0)` = 0
∴ x1 = x2
∴ f is a one-one function.
Let f(x) = 5x3 – 8 = y (say), y ∈ R
∴ x = `root(3)((y + 8)/5)`
∴ For every y ∈ R, there is some x ∈ R
∴ f is an onto function.
x = `root(3)((y + 8)/5)`
= f–1 (y)
∴ f–1 (x) = `root(3)((x + 8)/5)`
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